EGM 4313NAME:_____________________________Spring 2009Final ExamOpen book/Closed notes/No calculators or computers2 hour time limit(1)Solve the following heat equation problem:(29(29(29(29225,0,10,,20 ,,0105cos2uuuxtutu xxtxxπππ∂∂∂====++÷∂∂∂.First find the steady-state solution that meets the boundary conditions:(29(29(29(29(2901010201010ssssssssuxAxBduAdxuBuxxππππ=+=⇒==⇒=⇒=+(29(29(29(29(29(2922,,5,0,0,,0,,05cos2ssu x tuxx txttxtxxφφφφφ πφ=+∂∂∂⇒====÷∂∂∂The problem is greatly simplified if you notice that the initial condition on φmeets the boundary conditions and is in the form that one gets from separation of variables hence you should look for a solution of the form:(29(29(29(29(29(29(29(29(29(29(29252544254551255,5cos01, 5coscos2242255,5cos425,105cos2tttxxxx tT tTTtT tTtxT tex teT txu x txeφφπ---′=⇒== -÷÷÷′⇒= -⇒=⇒=÷⇒=++÷

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EGM 4313NAME:_____________________________Spring 2009Final Exam(4) The inverse of a Laplace transform is given by (291lim2ciTstTciTefs dsiπ+→∞-∫where cis chosen so that all the singular points of f(s) lie to the left of the line s= c(cis a real number). If (29asefss-=show that (290, 01,taftta<<=by appropriate integrals in the complex plane as shown. (1 point extra credit: Explain why the paths must be closed in the manner shown in each case.)(29ta sassteeess--=has a simple pole at s=0 with reside 1(a) t<a case:Here the integrand is analytic everywhere inside and on the contour hence the contour integral is 0. Therefore the inverse Laplace transform is 0 if the integrand vanishes on the arc as R→ ∞.

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